Question: $ \lim_{x\to -1}\sqrt{5-11x}=$
$\sqrt{5-11x}$ defines a square-root function. Square-root functions are continuous across their entire domain, and their domain is all real $x$ -values for which the expression within the square-root is non-negative. In other words, for any square-root function $q$ and any input $c$ in the domain of $q$ (except for its endpoint), we know that this equality holds: $\lim_{x\to c}q(x)=q(c)$ [What happens at the endpoint?] The input $x=-1$ is within the domain of $\sqrt{5-11x}$. Therefore, in order to find $ \lim_{x\to -1}\sqrt{5-11x}$, we can simply evaluate $\sqrt{5-11x}$ at $x=-1$. $\begin{aligned} &\phantom{=}\sqrt{5-11x} \\\\ &=\sqrt{5-11(-1)} \gray{\text{Substitute }x=-1} \\\\ &=\sqrt{16} \\\\ &=4 \end{aligned}$ In conclusion, $ \lim_{x\to -1}\sqrt{5-11x}=4$.